Lecture 8 dynamic programming

Algorithms Analysis lecture 8 Minimum and Maximum Alg + Dynamic Programming

Min and Max The minimum of a set of elements: The first order statistic i = 1 The maximum of a set of elements: The n-th order statistic i = n The median is the “halfway point” of the set i = (n+1)/2 , is unique when n is odd i =  (n+1)/2  = n/2 ( lower median ) and  (n+1)/2  = n/2+1 ( upper median ), when n is even

Finding Minimum or Maximum Alg.: MINIMUM (A, n) min ← A[1] for i ← 2 to n do if min > A[i] then min ← A[i] return min How many comparisons are needed? n – 1 : each element, except the minimum, must be compared to a smaller element at least once The same number of comparisons are needed to find the maximum The algorithm is optimal with respect to the number of comparisons performed

Simultaneous Min, Max Find min and max independently Use n – 1 comparisons for each  total of 2n – 2 At most 3n/2 comparisons are needed Process elements in pairs Maintain the minimum and maximum of elements seen so far Don’t compare each element to the minimum and maximum separately Compare the elements of a pair to each other Compare the larger element to the maximum so far, and compare the smaller element to the minimum so far This leads to only 3 comparisons for every 2 elements

Analysis of Simultaneous Min, Max Setting up initial values: n is odd: n is even: Total number of comparisons: n is odd: we do 3(n-1)/2 comparisons n is even: we do 1 initial comparison + 3(n-2)/2 more comparisons = 3n/2 - 2 comparisons set both min and max to the first element compare the first two elements , assign the smallest one to min and the largest one to max

Example: Simultaneous Min, Max n = 5 (odd), array A = {2, 7, 1, 3, 4} Set min = max = 2 Compare elements in pairs: 1 < 7  compare 1 with min and 7 with max  min = 1, max = 7 3 < 4  compare 3 with min and 4 with max  min = 1, max = 7 We performed: 3(n-1)/2 = 6 comparisons 3 comparisons 3 comparisons

Example: Simultaneous Min, Max n = 6 (even), array A = {2, 5, 3, 7, 1, 4} Compare 2 with 5: 2 < 5 Set min = 2, max = 5 Compare elements in pairs: 3 < 7  compare 3 with min and 7 with max  min = 2, max = 7 1 < 4  compare 1 with min and 4 with max  min = 1, max = 7 We performed: 3n/2 - 2 = 7 comparisons 1 comparison 3 comparisons 3 comparisons

Advanced Design and Analysis Techniques 􀂄 Covers important techniques for the design and analysis of efficient algorithms: such as dynamic programming , greedy algorithms .

Dynamic Programming Well known algorithm design techniques:. Divide-and-conquer algorithms Another strategy for designing algorithms is dynamic programming . Used when problem breaks down into recurring small subproblems Dynamic programming is typically applied to optimization problems . In such problem there can be many solutions . Each solution has a value, and we wish to find a solution with the optimal value.

Divide-and-conquer Divide-and-conquer method for algorithm design: Divide : If the input size is too large to deal with in a straightforward manner, divide the problem into two or more disjoint subproblems Conquer : conquer recursively to solve the subproblems Combine : Take the solutions to the subproblems and “merge” these solutions into a solution for the original problem

Dynamic programming Dynamic programming is a way of improving on inefficient divide-and-conquer algorithms. By “ inefficient ”, we mean that the same recursive call is made over and over . If same subproblem is solved several times , we can use table to store result of a subproblem the first time it is computed and thus never have to recompute it again. Dynamic programming is applicable when the subproblems are dependent , that is, when subproblems share subsubproblems. “ Programming” refers to a tabular method

Difference between DP and Divide-and-Conquer Using Divide-and-Conquer to solve these problems is inefficient because the same common subproblems have to be solved many times . DP will solve each of them once and their answers are stored in a table for future use.

Elements of Dynamic Programming (DP) DP is used to solve problems with the following characteristics : Simple subproblems We should be able to break the original problem to smaller subproblems that have the same structure Optimal substructure of the problems The optimal solution to the problem contains within optimal solutions to its subproblems . Overlapping sub-problems there exist some places where we solve the same subproblem more than once .

Steps to Designing a Dynamic Programming Algorithm Characterize optimal substructure 2. Recursively define the value of an optimal solution 3. Compute the value bottom up 4. (if needed) Construct an optimal solution

Fibonacci Numbers Fn= Fn-1+ Fn-2 n ≥ 2 F0 =0, F1 =1 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … Straightforward recursive procedure is slow ! Let’s draw the recursion tree

Fibonacci Numbers How many summations are there? Using Golden Ratio As you go farther and farther to the right in this sequence, the ratio of a term to the one before it will get closer and closer to the Golden Ratio. Our recursion tree has only 0s and 1s as leaves, thus we have 1.6 n summations Running time is exponential !

Fibonacci Numbers We can calculate Fn in linear time by remembering solutions to the solved subproblems – dynamic programming Compute solution in a bottom-up fashion In this case, only two values need to be remembered at any time

Ex1:Assembly-line scheduling 􀂄 Automobiles factory with two assembly lines. 􀂄 Each line has the same number “n” of stations. Numbered j = 1, 2, ..., n. 􀂅 We denote the jth station on line i (where i is 1 or 2) by Si,j . 􀂅 The jth station on line 1 (S1,j) performs the same function as the jth station on line 2 (S2,j ). 􀂅 The time required at each station varies, even between stations at the same position on the two different lines, as each assembly line has different technology. 􀂅 time required at station Si,j is (ai,j) . 􀂅 There is also an entry time (ei) for the chassis( هيكل ) to enter assembly line i and an exit time (xi) for the completed auto to exit assembly line i.

Ex1:Assembly-line scheduling (Time between adjacent stations are nearly 0).

Problem Definition Problem: Given all these costs, what stations should be chosen from line 1 and from line 2 for minimizing the total time for car assembly. “ Brute force” is to try all possibilities. requires to examine Omega(2 n ) possibilities Trying all 2 n subsets is infeasible when n is large. Simple example : 2 station  (2 n ) possibilities =4 start end

Step 1: Optimal Solution Structure 􀂄 optimal substructure : choosing the best path to Sij. The structure of the fastest way through the factory (from the starting point) The fastest possible way to get through Si ,1 ( i = 1, 2) Only one way: from entry starting point to Si ,1 take time is entry time ( ei)

Step 1: Optimal Solution Structure The fastest possible way to get through S i , j ( i = 1, 2) ( j = 2, 3, ..., n ). Two choices: Stay in the same line: Si , j -1  Si , j Time is Ti , j -1 + ai , j If the fastest way through Si , j is through Si , j -1, it must have taken a fastest way through Si , j -1 Transfer to other line: S 3- i , j -1  Si , j Time is T 3- i , j -1 + t 3- i , j -1 + ai , j Same as above

Step 1: Optimal Solution Structure An optimal solution to a problem finding the fastest way to get through Si , j contains within it an optimal solution to sub-problems finding the fastest way to get through either Si , j -1 or S 3- i , j -1 Fastest way from starting point to Si , j is either: The fastest way from starting point to Si , j -1 and then directly from Si , j -1 to Si , j or The fastest way from starting point to S 3- i , j -1 then a transfer from line 3- i to line i and finally to Si , j  Optimal Substructure.

Step 2: Recursive Solution Define the value of an optimal solution recursively in terms of the optimal solution to sub-problems Sub-problem here finding the fastest way through station j on both lines (i=1,2) Let fi [ j ] be the fastest possible time to go from starting point through Si , j The fastest time to go all the way through the factory : f * x 1 and x 2 are the exit times from lines 1 and 2, respectively

Step 2: Recursive Solution The fastest time to go through Si , j e 1 and e 2 are the entry times for lines 1 and 2

Step 2: Recursive Solution To help us keep track of how to construct an optimal solution, let us define li [ j ]: line # whose station j -1 is used in a fastest way through Si , j ( i = 1, 2, and j = 2, 3,..., n ) we avoid defining li [1] because no station precedes station 1 on either lines. We also define l *: the line whose station n is used in a fastest way through the entire factory

Step 2: Recursive Solution Using the values of l * and li [ j ] shown in Figure (b) in next slide, we would trace a fastest way through the factory shown in part (a) as follows The fastest total time comes from choosing stations Line 1: 1, 3, & 6 Line 2: 2, 4, & 5

Step 3: Optimal Solution Value

Step 4: Optimal Solution Constructing the fastest way through the factory

Lecture 8 dynamic programming

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Lecture 8 dynamic programming